Bertini's Theorem and singular divisors on a surface

I'm trying to understand the following:

Let $X$ be a projective, smooth surface over an algebraically closed field and $D$ a divisor on $X$. How can I see that $D$ is linear equivalent to the difference of two smooth curves on $X$ by Bertini's theorem?

Does proving this require the use of ample sheaves?

Any help would be very appreciated :)

Since $X$ is projective, there is an ample divisor $H$ on $X$. Then, as msteve says, $D+nH$ is ample for $n$ sufficiently large, but we can even have $D +nH$ and $nH$ both very ample for $n$ sufficiently large, since $D + kH$ is basepoint-free for large $k$ by amplitude, $lH$ is very ample for large $l$, and so $D + (k+l)H$ is very ample by this question.
Now we note that if $E$ is very ample on $X$, then almost all curves $E'$ in the complete linear system $\lvert E \rvert$ are smooth by embedding $X$ into some $\mathbf{P}^n$ using $E$, and then applying Bertini's Theorem (Hartshorne, Thm. II.8.18). Note this "Bertini's Theorem" holds in positive characteristic (but you still need $k$ to be infinite)!
Finally, choose $D' \in \lvert D + nH \rvert$ and $H' \in \lvert nH \rvert$ such that $D',H'$ are smooth as in the paragraph above. Then, $D \sim D' - H'$ is linearly equivalent to the difference of two smooth curves.
Take a hyperplane $H$ in $X$, then for $n$ sufficiently large, $D + nH$ is ample. Bertini's theorem says that there is a smooth effective divisor $E$ which is linearly equivalent to $D+nH$ (to use Bertini's theorem, I believe that our base field must be not only algebraically closed, but of characteristic zero). Then, $D$ is linearly equivalent to $E - nH$.
• Thank you :) Do you may be know where can I read the proof that $D + nH$ is ample for $n$ big enough? Is this easy to see? Unfortunately I am currently working with positive characteristic. – user1728 Jan 13 '16 at 21:56
• I think it follows from Serre vanishing that $D+nH$ is ample for $n \gg 0$. It's possible that there is a version of Bertini in positive characteristic that would work, but I do not know it. (Maybe there is something special in dimension $2$? Everything I've said above works for $X$ of any dimension.) – msteve Jan 13 '16 at 22:06